The polygon that finds the shape

Hang a string between two nails and load it with weights at five places. The string does not curve — it takes five kinks and six straight runs, because between the weights there is nothing to bend it.

That polygon is the funicular for those five loads, and it is fully determined by them plus one number. Change a weight and the polygon changes. Change how tightly the string is pulled and the polygon changes shape too, but only by scaling: the same set of directions, stretched.

One constant, and everything else follows #

Cut the string anywhere and look at the piece to one side. It carries the loads that hang from it, the pull at the far end, and the tension in the string at the cut. The string can carry force only along itself, so the tension’s direction is the string’s direction.

Now sum horizontally. Nothing horizontal is applied anywhere between the ends, so the horizontal component of the tension is the same at every station along the string. Call it $H$.

Sum vertically, and the vertical component at any station is the accumulated load between that station and the support. So the slope of the string at any point is

$$\frac{dy}{dx} = \frac{V(x)}{H},$$

which is one number divided by another that never changes. The polygon’s shape is therefore the running vertical sum, drawn — and since the running vertical sum is the shear diagram of an equivalent beam, the funicular polygon is a shear diagram integrated once, at a scale set by $H$.

That relationship is worth stating the other way round as well, because it is the more useful direction: the polygon’s vertical ordinate at any station is the beam’s bending moment there divided by $H$.

$$y(x) = \frac{M(x)}{H}.$$

Every funicular shape in existence is a moment diagram at a scale. The parabola of a uniformly loaded cable is the parabolic moment diagram of a uniformly loaded beam; the polygon above is the piecewise-linear moment diagram of a beam under five point loads. The shape that carries load without bending is the shape of the bending it avoids.

The pole, which turns it into a drawing #

Before there were equations there was a construction, and the construction is worth having because it explains what $H$ is doing.

Draw the loads end to end as a vertical line, to scale — this is the load line. Pick a point off to one side, at a horizontal distance from the load line representing $H$; this is the pole. Draw rays from the pole to every division on the load line.

Each ray now has the direction of one segment of the funicular polygon, because its slope is exactly one running vertical sum divided by the horizontal offset. Transfer the rays in order to the space diagram, each starting where the previous one ended, and the polygon appears. Two drawings, a straightedge, and no arithmetic at all — which is the whole claim graphic statics makes, demonstrated on the one problem where it is least replaceable.

The pole’s distance from the load line is $H$, and moving it is exactly the sag adjustment. Move the pole further out and the rays flatten, the polygon becomes shallow and the thrust grows. Move it in and the polygon deepens and the thrust falls. Every question about the trade between depth and force is answered by sliding one point across a sheet of paper.

Move the pole vertically instead and something different happens: the polygon tilts, which corresponds to the two supports being at different levels. The construction handles the case with no extra machinery, which is one of several reasons it survived so long after the algebra was available.

Sag against thrust, which is the only trade there is #

Since $y = M/H$ and $M$ is fixed by the loads, sag and thrust are strictly reciprocal. Doubling the sag halves the thrust; halving it doubles the thrust; and there is no arrangement in which both are small.

$$H = \frac{M_{\text{max}}}{\text{sag}}.$$

That reciprocal is worth watching happen rather than reading, because it is the one relationship in the essay with no approximation anywhere in it. The loads are unchanged, so the moment diagram they produce is unchanged, and the only free variable left is the number the whole diagram is divided by. Pulling the string tighter does not alter what it is carrying; it alters the scale at which the carrying is drawn.

Both numbers matter and they do not fall together. The thrust dropped by a third and the worst segment tension by only a seventh, because a segment carries $\sqrt{H^2 + V^2}$ and the vertical part is fixed by the loads. Sag buys thrust cheaply and member force dearly, which is why deepening a cable helps the anchorage far more than it helps the cable, and why the sag ratio is argued about by the people designing the ends rather than the middle.

Real cables sit at sag ratios between about a twelfth and a tenth of the span, and the reason is a compromise rather than an optimum. A deeper cable needs less material in the cable and taller towers to hold it; a shallower one needs less tower and enormously more anchorage. The Golden Gate’s main span sags about a tenth of its length; the Humber’s about a twelfth. That the numbers cluster so tightly across a century of very different bridges is the sign of a flat optimum, which is the same flatness that fixes span-to-depth ratios in beams and trusses.

Inverted, and used as a design tool #

The construction gives the shape that carries a stated set of loads in pure tension. Reverse every force and the same shape carries them in pure compression, which is Hooke’s principle and is exact.

Used forwards, the technique analyses. Used backwards it designs, and the backwards use is the interesting one: rather than choosing a shape and computing the bending in it, choose the loads and let the shape follow. Poleni did this in 1748 to assess the cracked dome of St Peter’s, hanging a chain loaded to represent the dome’s weight and checking that the inverted curve lay within the masonry. Gaudí spent a decade on a hanging model for the Colònia Güell chapel, photographing it and inverting the photographs.

What both were exploiting is that the model performs the optimisation itself. A hanging string cannot take a wrong shape — it has no bending stiffness with which to hold one — so the answer is found by the apparatus rather than computed by the designer. That is a rare property, and its modern equivalent is form-finding software in which the designer manipulates the force diagram and the program returns the geometry, which is the pole construction with a computer holding the straightedge.

There is not one polygon, there is a family #

Choosing $H$ chose a shape. Any other value of $H$ would have given a different shape, and every one of them is in equilibrium with the same loads. That is not an ambiguity in the method; it is a fact about the problem, and it turns out to be the most useful thing in this essay.

For a cable the ambiguity is resolved by the physical length of the string: a longer cable sags more, which fixes the sag, which fixes $H$. One shape, determined.

For an arch nothing resolves it. A masonry arch is a set of blocks that can transmit compression by any route the geometry permits, and the family of thrust lines in equilibrium with its loads is infinite. The arch is not obliged to pick one and tell anybody which.

What rescues the situation is a theorem rather than a measurement. The lower-bound theorem of plasticity says that if any thrust line can be drawn that is in equilibrium with the loads and stays inside the masonry, the arch will not collapse. It does not have to be the real one. It does not have to be identified, or unique, or even likely. It only has to exist.

So the assessment of a masonry arch becomes a search rather than a calculation: find one admissible member of the family. Heyman’s geometrical factor of safety makes the idea quantitative — it is the factor by which the arch’s thickness could be reduced before no admissible thrust line remains, which is a statement about how much room the family has to move in.

Two consequences follow that read as paradoxes and are not. A cracked arch has not failed; it has moved to a geometry in which some member of the family fits, and cracking is the mechanism by which it searched. And an arch whose real internal forces are entirely unknown can nevertheless be declared safe with confidence, because the theorem never asked what they were.

What the anchorages cost #

The thrust has to go somewhere, and where it goes is the largest single item in a suspension structure.

A cable’s pull at the top of a tower splits into a vertical component the tower carries down and a horizontal component that continues along the backstay to an anchorage. The anchorage has to resist that pull permanently, in tension, against nothing but its own weight and whatever the ground offers. On a major suspension bridge each anchorage is a block of concrete of the order of a hundred thousand tonnes, and it is quite normal for the two anchorages to cost more than the cable, the towers and the deck together.

That is the honest price of the funicular argument. The shape carries its load with no bending and no material wasted on resisting any, and it hands the entire difficulty to the two points at its ends. A structure whose members are perfectly efficient and whose foundations are enormous is the characteristic outcome, and it is why suspension bridges are built at crossings with rock close to the surface and are awkward at crossings without it.

There is an escape, and it has a cost of its own. A self-anchored suspension bridge takes the cable’s horizontal pull into the deck rather than into the ground, so the deck goes into compression and the anchorages disappear. The deck is then a compression member the length of the span, which must not buckle, and the whole structure has to be erected on falsework because the cable cannot be tensioned until there is a deck to pull against. A tied arch does the identical trade in reverse — thrust into a tie instead of into abutments — and pays for it with a member that must never fail.

The pattern is worth generalising, because it recurs whenever a structure is made efficient. Efficiency does not remove a difficulty; it relocates it, usually to a smaller number of larger components, and the question worth asking of any elegant structural form is where its problem has gone rather than whether it has one.

What happens when the load changes #

The polygon is funicular for one set of loads. Real structures see several, and this is where the elegance is paid for.

Those two polygons are the whole difficulty in one pair of pictures. A structure built to the first and asked to carry the second is being asked to change shape, and the gap between the axis it was built on and the axis the new loads want is bending in a member chosen to have none. The thrust hardly moved, so nothing at the anchorage announces the problem; what moved is the geometry, which is the one thing a built structure cannot do.

A suspension bridge shaped for its own dead weight is not funicular for a train on half of it. The cable cannot change shape without moving, and moving is what the deck resists — so the deck carries the difference as bending, and the stiffening truss exists for that purpose alone. On the Tacoma Narrows the stiffening was a shallow plate girder rather than a truss, and the deck’s response to loads its shape did not suit is the least interesting part of what happened to it and still a real part.

The same limitation constrains arches, where it is worse because compression is involved. A masonry arch’s thrust line moves when the load pattern moves, and if it moves outside the masonry the arch hinges. Four hinges make a mechanism, and that — not crushing — is how arches actually fall down.

The practical response in both cases is to make the structure heavy relative to the variable load, so that the dead-load funicular dominates. A masonry arch bridge with a deep fill over it is doing exactly this, and it is why the arrangement is so insensitive to where a load stands: the fill is not decorative, it is a way of ensuring the thrust line barely moves when a lorry crosses.

How wrong a polygon is as a curve #

The polygon and the parabola are usually treated as the same object with different resolutions, and the error between them has a closed form worth knowing.

Replace a uniform load $w$ over a span $L$ by $n$ equal point loads at the centres of $n$ panels. The two moment diagrams agree at the panel boundaries and differ inside them by the local parabola the discretisation threw away, $wp^2/8$ with $p = L/n$. Divide by $H$ and that is a sag:

$$\Delta = \frac{wL^2}{8H\,n^2} = \frac{\text{sag}}{n^2}$$

The polygon lies inside the curve by the sag divided by the square of the number of loads. Five loads is four per cent, ten is one, and twenty is a quarter of one.

Eighty-eight millimetres is the size of the whole discretisation error on this figure, and it is a useful thing to have a number for, because the polygon and the parabola are usually drawn on top of one another with no indication of which is which. At this scale the difference is a line width. It is also entirely one-sided: the polygon is always inside the curve, never outside, because a chord of a convex arc is.

That settles two practical questions at once and they point in opposite directions.

A suspension bridge does not care. Hangers at 10 m centres over a 1,000 m span give $n = 100$ and an error of one part in ten thousand — smaller than the cable’s own elastic stretch, and far smaller than the temperature movement. The main cable of a real bridge is a polygon and is drawn as a parabola with no apology needed.

A funicular roof does. Five or six purlins on a shaped truss is exactly the case where the four per cent is visible, and it is visible in the right place: the segments are straight and the designer who set them out from a parabola has built a shape that is not the funicular of the loads it carries. The correction is not to refine the drawing; it is to set the vertices out from the polygon, which the construction gives directly.

A load between the vertices is a different structure #

The polygon carries its loads in pure tension only because every load is at a vertex. Put one anywhere else and the member it lands on has to bend — and a funicular member is the worst possible thing to ask that of, because it was sized with no section modulus to spare.

Take the essay’s own figure: a span of 8 m, five loads, so segments about 1.5 m long carrying of the order of 40 kN. Suppose the member is a 50 mm round bar, $A = 1{,}963$ mm² and $W = 12{,}270$ mm³, and a 5 kN load arrives at the middle of one segment:

$$\sigma_a = \frac{40{,}000}{1{,}963} = 20\ \text{N/mm}^2, \qquad \sigma_b = \frac{5 \times 1{,}500/4}{12{,}270} = 155\ \text{N/mm}^2$$

The bending is nearly eight times the axial stress, from a load an eighth the size of the ones the shape was found for. The reason is entirely the section: a member chosen for axial force alone has a section modulus proportional to its diameter cubed over a force proportional to its diameter squared, so the smaller the axial member the worse it is at bending, and a funicular design makes every member as small as it can be.

Which turns a drawing rule into a structural one. The vertices are where the loads go, and nothing may be hung between them — not a service, not a light, not a walkway hanger, not a purlin moved 300 mm to clear something. On a cable that has no bending stiffness at all the same event simply produces a new vertex and a new shape; on a straight steel segment it produces a stress the member cannot take. The stiffer version of the funicular is the less forgiving one, which is the reverse of the usual expectation and the same reversal an imposed deformation produces everywhere else in this collection.

Where the model stops #

No bending stiffness. A real cable has a little and a real arch has a great deal — a section with a second moment of area — and the arch’s stiffness is what lets it tolerate a thrust line that has wandered. The funicular argument gives the shape at which bending is zero; it says nothing about how much bending the structure can survive when the shape is wrong.

One load case. As above, and it is the central limitation rather than a footnote.

Small deflections everywhere else, and large ones here. The usual first-order assumption does not hold for a cable at all: it changes shape substantially under load, so its geometry is an output rather than an input and the analysis is genuinely nonlinear. The polygon drawn here is the answer, not a step toward it.

Rigid supports. An arch’s thrust spreads its abutments, which flattens the arch, which raises the thrust. Several medieval arches record that history in their present geometry.

Self-weight not included as such. The construction treats the loads as given. A cable’s own weight is distributed along its length rather than along the horizontal, which is what makes the catenary a catenary, and the polygon method handles it only by chopping the cable into pieces and treating each piece’s weight as a point load — which converges, and is an approximation the drawing does not admit to.

The figures share one honest distortion. The polygon is drawn at a sag of roughly a quarter of its span, and the real thing is nearer a tenth. Since thrust and sag are reciprocal, every figure here understates the horizontal force by a factor of two or more — the shape is right and the consequence at the anchorages is drawn far smaller than it is.

The ladder from here #

Later rungs on this anchor: the pole diagram and Bow’s notation. Funicular polygons through three specified points. The catenary derived properly. Cable sag and the length problem. The thrust line in masonry and the middle-third rule. Heyman’s safe theorem and the geometrical factor of safety. Cable-stayed against suspended, and why the two behave differently. Prestressed cable nets. And computational form-finding, where the reciprocal diagram returns as an interactive tool.

Varignon published the funicular construction in 1725, having realised that the polygon of forces and the polygon of the string are two drawings of the same set of vectors. That the shape of a hanging chain and the diagram of the forces in it are reciprocal figures — each recoverable from the other — was established by Maxwell in 1864, and is the reason the construction works at all.